3.23.78 \(\int \frac {\sqrt {-x+x^4}}{-b+a x^6} \, dx\)

Optimal. Leaf size=173 \[ \frac {\sqrt {\sqrt {b} \left (\sqrt {a}-\sqrt {b}\right )} \tan ^{-1}\left (\frac {x \sqrt {x^4-x} \sqrt {\sqrt {a} \sqrt {b}-b}}{\sqrt {b} (x-1) \left (x^2+x+1\right )}\right )}{3 \sqrt {a} b}-\frac {\sqrt {-\left (\sqrt {b} \left (\sqrt {a}+\sqrt {b}\right )\right )} \tan ^{-1}\left (\frac {x \sqrt {x^4-x} \sqrt {-\sqrt {a} \sqrt {b}-b}}{\sqrt {b} (x-1) \left (x^2+x+1\right )}\right )}{3 \sqrt {a} b} \]

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Rubi [A]  time = 0.28, antiderivative size = 185, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2056, 1493, 1491, 1175, 402, 217, 206, 377, 205, 208} \begin {gather*} \frac {\sqrt {x^4-x} \sqrt {\sqrt {a}-\sqrt {b}} \tan ^{-1}\left (\frac {x^{3/2} \sqrt {\sqrt {a}-\sqrt {b}}}{\sqrt [4]{b} \sqrt {x^3-1}}\right )}{3 \sqrt {a} b^{3/4} \sqrt {x} \sqrt {x^3-1}}+\frac {\sqrt {x^4-x} \sqrt {\sqrt {a}+\sqrt {b}} \tanh ^{-1}\left (\frac {x^{3/2} \sqrt {\sqrt {a}+\sqrt {b}}}{\sqrt [4]{b} \sqrt {x^3-1}}\right )}{3 \sqrt {a} b^{3/4} \sqrt {x} \sqrt {x^3-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-x + x^4]/(-b + a*x^6),x]

[Out]

(Sqrt[Sqrt[a] - Sqrt[b]]*Sqrt[-x + x^4]*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*x^(3/2))/(b^(1/4)*Sqrt[-1 + x^3])])/(3
*Sqrt[a]*b^(3/4)*Sqrt[x]*Sqrt[-1 + x^3]) + (Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[-x + x^4]*ArcTanh[(Sqrt[Sqrt[a] + Sqr
t[b]]*x^(3/2))/(b^(1/4)*Sqrt[-1 + x^3])])/(3*Sqrt[a]*b^(3/4)*Sqrt[x]*Sqrt[-1 + x^3])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 1175

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2*r), In
t[(d + e*x^2)^q/(r - c*x^2), x], x] - Dist[c/(2*r), Int[(d + e*x^2)^q/(r + c*x^2), x], x]] /; FreeQ[{a, c, d,
e, q}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

Rule 1491

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m +
1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + c*x^((2*n)/k))^p, x], x, x^k], x] /; k !=
1] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m]

Rule 1493

Int[((f_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = D
enominator[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(k*n))/f)^q*(a + (c*x^(2*k*n))/f)^p, x], x, (f
*x)^(1/k)], x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && FractionQ[m] && IntegerQ[p
]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {\sqrt {-x+x^4}}{-b+a x^6} \, dx &=\frac {\sqrt {-x+x^4} \int \frac {\sqrt {x} \sqrt {-1+x^3}}{-b+a x^6} \, dx}{\sqrt {x} \sqrt {-1+x^3}}\\ &=\frac {\left (2 \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt {-1+x^6}}{-b+a x^{12}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt {-1+x^3}}\\ &=\frac {\left (2 \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2}}{-b+a x^4} \, dx,x,x^{3/2}\right )}{3 \sqrt {x} \sqrt {-1+x^3}}\\ &=-\frac {\left (\sqrt {a} \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2}}{\sqrt {a} \sqrt {b}-a x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}-\frac {\left (\sqrt {a} \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {-1+x^2}}{\sqrt {a} \sqrt {b}+a x^2} \, dx,x,x^{3/2}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}\\ &=-\frac {\left (\sqrt {a} \left (-1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (\sqrt {a} \sqrt {b}-a x^2\right )} \, dx,x,x^{3/2}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}+\frac {\left (\sqrt {a} \left (1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2} \left (\sqrt {a} \sqrt {b}+a x^2\right )} \, dx,x,x^{3/2}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}\\ &=-\frac {\left (\sqrt {a} \left (-1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a} \sqrt {b}-\left (-a+\sqrt {a} \sqrt {b}\right ) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}+\frac {\left (\sqrt {a} \left (1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \sqrt {-x+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a} \sqrt {b}-\left (a+\sqrt {a} \sqrt {b}\right ) x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {-1+x^3}}\right )}{3 \sqrt {b} \sqrt {x} \sqrt {-1+x^3}}\\ &=\frac {\sqrt {\sqrt {a}-\sqrt {b}} \sqrt {-x+x^4} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} x^{3/2}}{\sqrt [4]{b} \sqrt {-1+x^3}}\right )}{3 \sqrt {a} b^{3/4} \sqrt {x} \sqrt {-1+x^3}}+\frac {\sqrt {\sqrt {a}+\sqrt {b}} \sqrt {-x+x^4} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} x^{3/2}}{\sqrt [4]{b} \sqrt {-1+x^3}}\right )}{3 \sqrt {a} b^{3/4} \sqrt {x} \sqrt {-1+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 280, normalized size = 1.62 \begin {gather*} \frac {\sqrt {x} \sqrt {x^3-1} \left (\sqrt {\sqrt {a}-\sqrt {b}} \left (\tan ^{-1}\left (\frac {\sqrt [4]{a}-\sqrt [4]{b} x^{3/2}}{\sqrt {x^3-1} \sqrt {\sqrt {a}-\sqrt {b}}}\right )-\tan ^{-1}\left (\frac {\sqrt [4]{a}+\sqrt [4]{b} x^{3/2}}{\sqrt {x^3-1} \sqrt {\sqrt {a}-\sqrt {b}}}\right )\right )-\sqrt {\sqrt {a}+\sqrt {b}} \tanh ^{-1}\left (\frac {-\sqrt [4]{b} x^{3/2}+i \sqrt [4]{a}}{\sqrt {x^3-1} \sqrt {\sqrt {a}+\sqrt {b}}}\right )+\sqrt {\sqrt {a}+\sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x^{3/2}+i \sqrt [4]{a}}{\sqrt {x^3-1} \sqrt {\sqrt {a}+\sqrt {b}}}\right )\right )}{6 \sqrt {a} b^{3/4} \sqrt {x \left (x^3-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-x + x^4]/(-b + a*x^6),x]

[Out]

(Sqrt[x]*Sqrt[-1 + x^3]*(Sqrt[Sqrt[a] - Sqrt[b]]*(ArcTan[(a^(1/4) - b^(1/4)*x^(3/2))/(Sqrt[Sqrt[a] - Sqrt[b]]*
Sqrt[-1 + x^3])] - ArcTan[(a^(1/4) + b^(1/4)*x^(3/2))/(Sqrt[Sqrt[a] - Sqrt[b]]*Sqrt[-1 + x^3])]) - Sqrt[Sqrt[a
] + Sqrt[b]]*ArcTanh[(I*a^(1/4) - b^(1/4)*x^(3/2))/(Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[-1 + x^3])] + Sqrt[Sqrt[a] +
Sqrt[b]]*ArcTanh[(I*a^(1/4) + b^(1/4)*x^(3/2))/(Sqrt[Sqrt[a] + Sqrt[b]]*Sqrt[-1 + x^3])]))/(6*Sqrt[a]*b^(3/4)*
Sqrt[x*(-1 + x^3)])

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IntegrateAlgebraic [A]  time = 0.79, size = 173, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {-\left (\left (\sqrt {a}+\sqrt {b}\right ) \sqrt {b}\right )} \tan ^{-1}\left (\frac {\sqrt {-\sqrt {a} \sqrt {b}-b} x \sqrt {-x+x^4}}{\sqrt {b} (-1+x) \left (1+x+x^2\right )}\right )}{3 \sqrt {a} b}+\frac {\sqrt {\left (\sqrt {a}-\sqrt {b}\right ) \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a} \sqrt {b}-b} x \sqrt {-x+x^4}}{\sqrt {b} (-1+x) \left (1+x+x^2\right )}\right )}{3 \sqrt {a} b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[-x + x^4]/(-b + a*x^6),x]

[Out]

-1/3*(Sqrt[-((Sqrt[a] + Sqrt[b])*Sqrt[b])]*ArcTan[(Sqrt[-(Sqrt[a]*Sqrt[b]) - b]*x*Sqrt[-x + x^4])/(Sqrt[b]*(-1
+ x)*(1 + x + x^2))])/(Sqrt[a]*b) + (Sqrt[(Sqrt[a] - Sqrt[b])*Sqrt[b]]*ArcTan[(Sqrt[Sqrt[a]*Sqrt[b] - b]*x*Sq
rt[-x + x^4])/(Sqrt[b]*(-1 + x)*(1 + x + x^2))])/(3*Sqrt[a]*b)

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fricas [B]  time = 1.07, size = 1083, normalized size = 6.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/2)/(a*x^6-b),x, algorithm="fricas")

[Out]

-1/12*sqrt(-(a*b*sqrt(1/(a*b^3)) - 1)/(a*b))*log((2*((a^2 - 3*a*b + 4*b^2)*x^4 + 2*(a*b - 2*b^2)*x + (2*(a^2*b
^2 - 2*a*b^3)*x^4 + (a^2*b^2 - 3*a*b^3 + 4*b^4)*x)*sqrt(1/(a*b^3)))*sqrt(x^4 - x) + ((a^2*b - 4*a*b^2)*x^6 + 2
*(a^2*b - 3*a*b^2 + 4*b^3)*x^3 + 3*a*b^2 - 4*b^3 + ((a^3*b^2 - 6*a^2*b^3 + 8*a*b^4)*x^6 + a^2*b^3 + 4*(a^2*b^3
- 2*a*b^4)*x^3)*sqrt(1/(a*b^3)))*sqrt(-(a*b*sqrt(1/(a*b^3)) - 1)/(a*b)))/(a*x^6 - b)) + 1/12*sqrt(-(a*b*sqrt(
1/(a*b^3)) - 1)/(a*b))*log((2*((a^2 - 3*a*b + 4*b^2)*x^4 + 2*(a*b - 2*b^2)*x + (2*(a^2*b^2 - 2*a*b^3)*x^4 + (a
^2*b^2 - 3*a*b^3 + 4*b^4)*x)*sqrt(1/(a*b^3)))*sqrt(x^4 - x) - ((a^2*b - 4*a*b^2)*x^6 + 2*(a^2*b - 3*a*b^2 + 4*
b^3)*x^3 + 3*a*b^2 - 4*b^3 + ((a^3*b^2 - 6*a^2*b^3 + 8*a*b^4)*x^6 + a^2*b^3 + 4*(a^2*b^3 - 2*a*b^4)*x^3)*sqrt(
1/(a*b^3)))*sqrt(-(a*b*sqrt(1/(a*b^3)) - 1)/(a*b)))/(a*x^6 - b)) - 1/12*sqrt((a*b*sqrt(1/(a*b^3)) + 1)/(a*b))*
log((2*((a^2 - 3*a*b + 4*b^2)*x^4 + 2*(a*b - 2*b^2)*x - (2*(a^2*b^2 - 2*a*b^3)*x^4 + (a^2*b^2 - 3*a*b^3 + 4*b^
4)*x)*sqrt(1/(a*b^3)))*sqrt(x^4 - x) + ((a^2*b - 4*a*b^2)*x^6 + 2*(a^2*b - 3*a*b^2 + 4*b^3)*x^3 + 3*a*b^2 - 4*
b^3 - ((a^3*b^2 - 6*a^2*b^3 + 8*a*b^4)*x^6 + a^2*b^3 + 4*(a^2*b^3 - 2*a*b^4)*x^3)*sqrt(1/(a*b^3)))*sqrt((a*b*s
qrt(1/(a*b^3)) + 1)/(a*b)))/(a*x^6 - b)) + 1/12*sqrt((a*b*sqrt(1/(a*b^3)) + 1)/(a*b))*log((2*((a^2 - 3*a*b + 4
*b^2)*x^4 + 2*(a*b - 2*b^2)*x - (2*(a^2*b^2 - 2*a*b^3)*x^4 + (a^2*b^2 - 3*a*b^3 + 4*b^4)*x)*sqrt(1/(a*b^3)))*s
qrt(x^4 - x) - ((a^2*b - 4*a*b^2)*x^6 + 2*(a^2*b - 3*a*b^2 + 4*b^3)*x^3 + 3*a*b^2 - 4*b^3 - ((a^3*b^2 - 6*a^2*
b^3 + 8*a*b^4)*x^6 + a^2*b^3 + 4*(a^2*b^3 - 2*a*b^4)*x^3)*sqrt(1/(a*b^3)))*sqrt((a*b*sqrt(1/(a*b^3)) + 1)/(a*b
)))/(a*x^6 - b))

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giac [A]  time = 1.10, size = 207, normalized size = 1.20 \begin {gather*} \frac {{\left (4 \, \sqrt {a b} \sqrt {-b^{2} - \sqrt {a b} b} a + 5 \, \sqrt {a b} \sqrt {-b^{2} - \sqrt {a b} b} b\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {-\frac {1}{x^{3}} + 1}}{\sqrt {-\frac {b + \sqrt {{\left (a - b\right )} b + b^{2}}}{b}}}\right )}{3 \, {\left (4 \, a^{2} b^{3} + 5 \, a b^{4}\right )}} - \frac {{\left (4 \, \sqrt {a b} \sqrt {-b^{2} + \sqrt {a b} b} a + 5 \, \sqrt {a b} \sqrt {-b^{2} + \sqrt {a b} b} b\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {-\frac {1}{x^{3}} + 1}}{\sqrt {-\frac {b - \sqrt {{\left (a - b\right )} b + b^{2}}}{b}}}\right )}{3 \, {\left (4 \, a^{2} b^{3} + 5 \, a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/2)/(a*x^6-b),x, algorithm="giac")

[Out]

1/3*(4*sqrt(a*b)*sqrt(-b^2 - sqrt(a*b)*b)*a + 5*sqrt(a*b)*sqrt(-b^2 - sqrt(a*b)*b)*b)*abs(b)*arctan(sqrt(-1/x^
3 + 1)/sqrt(-(b + sqrt((a - b)*b + b^2))/b))/(4*a^2*b^3 + 5*a*b^4) - 1/3*(4*sqrt(a*b)*sqrt(-b^2 + sqrt(a*b)*b)
*a + 5*sqrt(a*b)*sqrt(-b^2 + sqrt(a*b)*b)*b)*abs(b)*arctan(sqrt(-1/x^3 + 1)/sqrt(-(b - sqrt((a - b)*b + b^2))/
b))/(4*a^2*b^3 + 5*a*b^4)

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maple [C]  time = 0.54, size = 365, normalized size = 2.11

method	result	size
default	\(-\frac {\sqrt {4}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (a \,\textit {\_Z}^{6}-b \right )}{\sum }\frac {\left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+1\right ) \left (-1+x \right )^{2} \left (\underline {\hspace {1.25 ex}}\alpha ^{5}+\underline {\hspace {1.25 ex}}\alpha ^{4}+\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}+\underline {\hspace {1.25 ex}}\alpha +1\right ) \left (1-i \sqrt {3}\right ) \sqrt {\frac {x \left (-3+i \sqrt {3}\right )}{\left (-1+x \right ) \left (-1+i \sqrt {3}\right )}}\, \sqrt {\frac {i \sqrt {3}+2 x +1}{\left (-1+x \right ) \left (-1-i \sqrt {3}\right )}}\, \sqrt {\frac {1+2 x -i \sqrt {3}}{\left (-1+x \right ) \left (-1+i \sqrt {3}\right )}}\, \left (\EllipticF \left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )-\frac {\underline {\hspace {1.25 ex}}\alpha ^{5} a \EllipticPi \left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \frac {i \underline {\hspace {1.25 ex}}\alpha ^{5} \sqrt {3}\, a -3 \underline {\hspace {1.25 ex}}\alpha ^{5} a -i \sqrt {3}\, b +3 b}{6 b}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )}{b}\right )}{\underline {\hspace {1.25 ex}}\alpha ^{4} \left (a -b \right ) \left (-3+i \sqrt {3}\right ) \sqrt {x \left (-1+x \right ) \left (i \sqrt {3}+2 x +1\right ) \left (1+2 x -i \sqrt {3}\right )}}\right )}{3}\)	\(365\)
elliptic	\(-\frac {\sqrt {4}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (a \,\textit {\_Z}^{6}-b \right )}{\sum }\frac {\left (-\underline {\hspace {1.25 ex}}\alpha ^{3}+1\right ) \left (-1+x \right )^{2} \left (\underline {\hspace {1.25 ex}}\alpha ^{5}+\underline {\hspace {1.25 ex}}\alpha ^{4}+\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}+\underline {\hspace {1.25 ex}}\alpha +1\right ) \left (1-i \sqrt {3}\right ) \sqrt {\frac {x \left (-3+i \sqrt {3}\right )}{\left (-1+x \right ) \left (-1+i \sqrt {3}\right )}}\, \sqrt {\frac {i \sqrt {3}+2 x +1}{\left (-1+x \right ) \left (-1-i \sqrt {3}\right )}}\, \sqrt {\frac {1+2 x -i \sqrt {3}}{\left (-1+x \right ) \left (-1+i \sqrt {3}\right )}}\, \left (\EllipticF \left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )-\frac {\underline {\hspace {1.25 ex}}\alpha ^{5} a \EllipticPi \left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-1+x \right )}}, \frac {i \underline {\hspace {1.25 ex}}\alpha ^{5} \sqrt {3}\, a -3 \underline {\hspace {1.25 ex}}\alpha ^{5} a -i \sqrt {3}\, b +3 b}{6 b}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )}{b}\right )}{\underline {\hspace {1.25 ex}}\alpha ^{4} \left (a -b \right ) \left (-3+i \sqrt {3}\right ) \sqrt {x \left (-1+x \right ) \left (i \sqrt {3}+2 x +1\right ) \left (1+2 x -i \sqrt {3}\right )}}\right )}{3}\)	\(365\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-x)^(1/2)/(a*x^6-b),x,method=_RETURNVERBOSE)

[Out]

-1/3*4^(1/2)*sum((-_alpha^3+1)/_alpha^4*(-1+x)^2*(_alpha^5+_alpha^4+_alpha^3+_alpha^2+_alpha+1)/(a-b)*(1-I*3^(
1/2))*(x/(-1+x)*(-3+I*3^(1/2))/(-1+I*3^(1/2)))^(1/2)*(1/(-1+x)*(I*3^(1/2)+2*x+1)/(-1-I*3^(1/2)))^(1/2)*(1/(-1+
x)*(1+2*x-I*3^(1/2))/(-1+I*3^(1/2)))^(1/2)/(-3+I*3^(1/2))/(x*(-1+x)*(I*3^(1/2)+2*x+1)*(1+2*x-I*3^(1/2)))^(1/2)
*(EllipticF(((-3/2+1/2*I*3^(1/2))*x/(-1/2+1/2*I*3^(1/2))/(-1+x))^(1/2),((3/2+1/2*I*3^(1/2))*(1/2-1/2*I*3^(1/2)
)/(1/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))-_alpha^5*a/b*EllipticPi(((-3/2+1/2*I*3^(1/2))*x/(-1/2+1/2*I*
3^(1/2))/(-1+x))^(1/2),1/6*(I*_alpha^5*3^(1/2)*a-3*_alpha^5*a-I*3^(1/2)*b+3*b)/b,((3/2+1/2*I*3^(1/2))*(1/2-1/2
*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))),_alpha=RootOf(_Z^6*a-b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{4} - x}}{a x^{6} - b}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/2)/(a*x^6-b),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 - x)/(a*x^6 - b), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\sqrt {x^4-x}}{b-a\,x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4 - x)^(1/2)/(b - a*x^6),x)

[Out]

-int((x^4 - x)^(1/2)/(b - a*x^6), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )}}{a x^{6} - b}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-x)**(1/2)/(a*x**6-b),x)

[Out]

Integral(sqrt(x*(x - 1)*(x**2 + x + 1))/(a*x**6 - b), x)

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